Solving an Integral Problem: e^{-x^2}

[Jbreezy]

Homework Statement

Hey, just stuck on an integral. I actually gave up after a while and typed it in wolfram and got something that is beyond me. Just looking for some clarity.

Homework Equations

<br /> I = \int e^{-x^2} \, dx<br />

The Attempt at a Solution

The only way I know to tackle e is to make a substitution for the variable being raised to a power. I tried to make u = x^2 but that doesn't go anywhere. The problem clearly lies in the fact that the exponent on e is being raised to a power it's self. I don't know how to deal with this. If you pop that sucker in on wolfram you get something I have never seen ...called erf(x) ...not a clue. So I'm looking for an explanation of this. Please and thanks.

 

[Ray Vickson]

Homework Statement

Hey, just stuck on an integral. I actually gave up after a while and typed it in wolfram and got something that is beyond me. Just looking for some clarity.

Homework Equations

<br /> I = \int e^{-x^2} \, dx<br />

The Attempt at a Solution

The only way I know to tackle e is to make a substitution for the variable being raised to a power. I tried to make u = x^2 but that doesn't go anywhere. The problem clearly lies in the fact that the exponent on e is being raised to a power it's self. I don't know how to deal with this. If you pop that sucker in on wolfram you get something I have never seen ...called erf(x) ...not a clue. So I'm looking for an explanation of this. Please and thanks.

The function $e^{-x^2}$ does not have an "elementary" anti-derivative; that means that it is impossible to write a *finite* formula for $I$ in terms of things like sums, products, reciprocals, powers, roots, exponentials, logarithms, trig functions, etc. Even if you allow formulas 10 million pages in length, you still cannot do it! BTW: it is NOT just that nobody has been smart enough to figure out how to write such a formula; rather, it has been proven to be impossible.

That is why functions like $\text{erf}(x)$have been invented; they allow us to express such integrals in a compact form, and numerical computations using them are not much more complicated than with $\log(x)$ or $\cos(x)$, etc.

 

[Jbreezy]

Whoa. Cool. So how do you do integrals when you see something like I have written above. Can you elaborate on how to deal with them? Thanks amig

 

[SteamKing]

That's the point. These integrals are evaluated numerically or they are left in their original form. You can't do anything else with them.

 

[Curious3141]

Homework Statement

Hey, just stuck on an integral. I actually gave up after a while and typed it in wolfram and got something that is beyond me. Just looking for some clarity.

Homework Equations

<br /> I = \int e^{-x^2} \, dx<br />

The Attempt at a Solution

The only way I know to tackle e is to make a substitution for the variable being raised to a power. I tried to make u = x^2 but that doesn't go anywhere. The problem clearly lies in the fact that the exponent on e is being raised to a power it's self. I don't know how to deal with this. If you pop that sucker in on wolfram you get something I have never seen ...called erf(x) ...not a clue. So I'm looking for an explanation of this. Please and thanks.

Just to add to Ray's post, $\text{erf}$ is called the "error function".

You need to invoke $\text{erf}(x)$ when doing the indefinite integral. However, if you wanted to evaluate the definite integral between certain special bounds, say between 0 and ∞, there is a way to do so exactly without needing $\text{erf}$, but the mathematics is slightly involved.

 

[Jbreezy]

Well the question was actually asking me to state if the limit converged or not between - infinity and infinity.
So how does that work? If you don't mind or maybe someone else knows.
Thank you.

 

[LCKurtz]

Well the question was actually asking me to state if the limit converged or not between - infinity and infinity.
So how does that work? If you don't mind or maybe someone else knows.
Thank you.

The integrand is an even function and $\int_{1}^{\infty}e^{-x^2}\, dx \le \int_{1}^{\infty}e^{-x}\, dx$

 

[Curious3141]

Well the question was actually asking me to state if the limit converged or not between - infinity and infinity.
So how does that work? If you don't mind or maybe someone else knows.
Thank you.

The method I alluded to allows you to actually calculate that integral. You don't need to do that just to prove that the integral is finite. LCKurtz's observation works fine.

 

[Jbreezy]

Lckrutz, What is it about it being even? I graphed them, I see but when you first looked what did you think?
P.S. I have a vector question. In another thread in a moment.

 

[LCKurtz]

Lckrutz, What is it about it being even? I graphed them, I see but when you first looked what did you think?
P.S. I have a vector question. In another thread in a moment.

For an even function $f(x)$, the graph is symmetric about the y axis. So if the part where $x>0$ converges, so does the part where $x<0$, and$$
\int_{-\infty}^{\infty} f(x)\, dx = 2\int_0^\infty f(x)\, dx$$

 

[Jbreezy]

OK thanks